Important numerical

Important numerical related to the hypothesis.

1. A sample of 400 male students is found to have a mean height of 171.38cm . Can it be reasonable regarded as a sample from a large population with mean height 171.17cm and standard deviation 3.30cm?.

$$Solution$$.

$$Sample\,size(n)=400$$

$$Sample\,mean \,\overline{X}=171.38cm$$

$$Population\,mean\,(µ)=171.17cm$$

$$Population\,standard\,deviation(σ)=3.3cm$$

$$H_0=171.7cm.That\,is\,the\,sample\,is\,form\,a\,normal\,population$$

$$H_1≠171.17c.m. (two\,tailed\,test). That\,is\,the\,sample\,is\,not\,form\,a\,normal\,population$$

$$Test \,statistic:Under\,H_0\,the\,test\,statistic\,is$$

$$Z=\frac{\overline{X}-µ}{\frac{σ}{\sqrt{n}}}$$

$$Z=\frac{171.38-171.17}{\frac{3.30}{\sqrt{400}}}$$

$$=1.27$$

$$z_{tab}\,at\,5\%\,level\,of\,significance\,for\,two\,tail\,test=1.96$$

Decision: Since, Z_calc<Z_tab at 5% level of significance for two tail tset, it is not significance and H₀ is accepted which means that the sample is from normal population.

2. A company has been producing steel tubes of the mean linear diameter of 2.00 cm. a sample of 10 tubes gives an inner diameter a mean diameter of 2.01 cms. is the difference in the values of mean significant.

$$Solution$$

$$Population\,mean(µ)=2cm$$

$$Sample\,size(n)=10$$

$$Sample\,mean \,\overline{X}=2.01cm$$

$$Varience(S^2)=0.004sq.cm$$

$$=S=0.0632cm$$

$$H_0=2cm. That\,is\,,there\,is\,no\,significant\,difference\,between\,sample\,mean\,and\,population\,mean$$

$$H_1≠2c.m.(two\,tailed\,test).there\,is\,significant\,difference\,between\,sample\,mean\,and\,population\,mean$$

$$Test\,statistic:UnderH_0\,the\,test\,statistics\,is$$

$$Z=\frac{\overline{X}-µ}{\frac{σ}{\sqrt{n}}}$$

$$Z=\frac{\overline{X}-µ}{\frac{s}{\sqrt{n}}}$$

$$Z=\frac{2.01-2}{\frac{0.0632}{\sqrt{10-1}}}$$

$$=0.475$$

$$t_{tab}\,value\,of\,t'\,for\,n-1\,=10-1=9d.f\,at\,5\%\,level\,of\,significance\,for\,two\,tail\,test\,is\,2.262$$

Decision: Since, tcal<ttab at 5% level of significance and H0 is accepted which means that the sample is form normal population.

Three hundred digits were chosen at random from a set of tables. The frequencies of the digits were are follows.

Using chi-square test (Χ²) assess the hypothesis that the digits were distributed in equal number in that table.

Solution.

H₀: There is no significant difference between the observed and expected frequency. In other words, the digits were distributed in equal number in the table.

H₁: There is the significant difference between the observed and expected frequency. In other words, the digits were not distributed in equal number in the table.

Test statistic.

Under H₀ the test statistic is.

$$Χ^2=∑\frac{(O-E)^2}{E}$$

$$Where\,O=Observed\,frequency$$

$$E=Expected\,frequency$$

Calculation of chi square.

$$Here,E=\frac{∑O}{n}$$

$$Here,E=\frac{300}{10}=30$$

$$Χ^2=∑\frac{(O-E)^2}{E}=2.8664$$

$$χ^2\,tab\,at\,n-1=10-1=9d.f for 5\%\,level\,of\,significance\,is\,16.92$$

Decision.

Sinceχ2 _cal <χ2 _tab at 5% level of significance for 9 d.f. It is not significant and H0 is accepted, which means that there is no significance difference between the observed and expected frequency.

In other words, the digits were not distributed in equal number in the table.

Reference

Kerlinger, F.N. Foundation of Behavioural Research. New Delhi: Surjeet Publication, 2000.

Kothari, C.R. Research Methodology. India: Vishwa Prakashan, 1990.

Singh, M.L. and J.M Singh. Understanding Research Methodology. 1998.

Singh, Mrigendra Lal. Understanding Research Methodology. Nepal: National Book centre, 2013.